2015多校联赛 ——HDU5288(数学)
2015-07-22 17:50
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OO’s Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 449 Accepted Submission(s): 158
Problem Description
OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know
∑i=1n∑j=inf(i,j) mod (10^9+7).
Input
There are multiple test cases. Please process till EOF.
In each test case:
First line: an integer n(n<=10^5) indicating the size of array
Second line:contain n numbers ai(0<ai<=10000)
Output
For each tests: ouput a line contain a number ans.
Sample Input
5 1 2 3 4 5
Sample Output
23
Source
2015 Multi-University Training Contest 1
求在[a , b] 中有多少个i 满足找不到另一个数j 使,a[i] % a[j] = 0
用L[],R[]分别记录左右离起最近的因子,ans+=(R[i]-i+1)*(i-L[i]+1)/2 %MOD
//自己并没有想到怎么做 (╯▽╰)
#include<cstdio> #include<cstring> #include<iostream> using namespace std ; const int maxn = 100010 ; const int mod = 1e9+7 ; int a[maxn]; long long last[maxn] ; long long pre[maxn] ; long long l[maxn] ; long long r[maxn] ; int main() { //freopen("1001.txt" ,"r" ,stdin) ; int n; while(~scanf("%d",&n)) { memset(pre,0,sizeof(pre)); memset(last,0,sizeof(last)); for(int i = 1;i <= n;i++) { l[i] = 1; r[i] = n; scanf("%d",&a[i]); for(int j = a[i];j < 10001;j+=a[i]) { if(pre[j] && r[pre[j]] == n) r[pre[j]] = i-1; pre[a[i]] = i; } } for(int i = n;i >=1;i--) { for(int j = a[i];j < 10001;j+=a[i]) { if(last[j] && l[last[j]]==1) l[last[j]] = i+1; last[a[i]] = i; } } long long ans = 0; for(long long int i = 1;i <= n;i++) ans= (ans+(((i-l[i]+1)%mod)*((r[i] - i +1)%mod))%mod)%mod; printf("%I64d\n",ans); } return 0 ; }
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